Home
Class 12
PHYSICS
Dimension of epsilon(0) are...

Dimension of `epsilon_(0)` are

A

`[M^(-1)L^(-3)T^(4)A^(2)]`

B

`[M^(-1)L^(-3)T^(2)A^(4)]`

C

`[ML^(3)T^(-4)A^(-2)]`

D

`[M^(-1)L^(-3)A^(2)T^(2)]`

Text Solution

Verified by Experts

The correct Answer is:
A
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROSTATICS

    MOTION|Exercise Exercise -3 Section (A)|41 Videos

  • ELECTROSTATICS

    MOTION|Exercise Exercise -3 Section (B) Previous Year Problems | JEE Main|30 Videos

  • ELECTROSTATICS

    MOTION|Exercise Exercise -1 (Objective Problems | NEET)|105 Videos

  • ELECTRONICS - SEMI CONDUCTOR

    MOTION|Exercise EXERCISE - 3|29 Videos

  • ELECTROSTATICS - I

    MOTION|Exercise EXERCISE - 4 (Level -II) PREVIOUS YEAR - JEE ADVANCED|13 Videos

Similar Questions

Explore conceptually related problems

Dimensions of epsilon_(0) are M^(-1)L^(-3)T^(4)A^(2) M^(0)L^(-3)T^(3)A^(3) M^(-1)L^(-3)T^(3)A M^(-1)L^(-3)TA^(2)

Find the dimensions of Gepsilon_(0) (G = Universal Gravitational constant, epsilon_(0)= permitivity in vaccum).

Knowledge Check

  • Dimensions of epsilon_(0) are

    A
    `[M^(-1)L^(3)T^(4)A^(2)]`
    B
    `[M^(0)L^(-3)T^(3)A^(3)]`
    C
    `[M^(-1)L^(-3)T^(3)A]`
    D
    `[M^(-1)L^(-3)TA^(2)]`

  • epsilon_(0)E^(2) has the dimensions of ( epsilon_(0)= permittivity of free space, E= electric field) Here k= Boltzmann consant T= absolute temperature R= universal gas constant.

    A
    Pressure
    B
    `kT`
    C
    `R//T`
    D
    `RT`

  • Consider a parallel plate capacitor, of capacitance C, plate area A and plate separation d. It is being charged by using a battery. The quantity epsilon_(0)(d phi_(E))/(dt) has the dimensions of [ epsilon_(0) =permittivity of free space, phi_(E) is the electric flux, dt = time]

    A
    e.m.f.
    B
    current
    C
    resistance
    D
    frequency

    • Similar Questions

    Explore conceptually related problems

    What are the dimensions of mu_(0)epsilon_(0) ? Here mu_(0)= magnetic permeability in vacuum, epsilon_(0)= electric permittivity in vacuum

    The dimensions of (1)/(epsilon_(0))(e^(2))/(hc) are

    The dimensions of epsilon_(0)mu_(0) are

    Find the dimension of epsilon_(0)E^(2) , where epsilon_(0) is permittivity of free space and E is the electricity field

    In SI units, the dimensions of sqrt((epsilon_(0))/(mu_(0))) is

    Home
    Profile